cosA +tanAsinA = secA
prove that the left hand side equals the right hand side.
2007-06-10 19:05:41 · 14 answers · asked by stacey_b 2 in Science & Mathematics ➔ Mathematics
cosA +tanAsinA = secA
cosA+(sinA/cosA)sinA = secA
cosA + (sinA)^2 / cosA = secA
(cosA)^2 /cosA + (sinA)^2 / cosA = secA
[(cosA)^2 + (sinA)^2]/cosA = secA
1 / cosA = secA
secA = secA
2007-06-10 19:11:09 · answer #1 · answered by espms290 4 · 1⤊ 0⤋
Divide both sides by sec A
Since 1/sec A = cos A, the first term of the divided equation will be cos^2 A.
Since tan A = sin A/cos A and cos A * sec A = 1, the sectond term of the divided equation will be sin^2 A.
The right hand side term is 1.
The result is an identity, which is proof.
2007-06-11 02:16:07 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
cosA + (sinA / cosA) sinA = secA
cosA + (sin^2 A / cosA) =
(cos^2 A / cos A) + (sin^2 A / cosA) =
(cos^2 A + sin^2 A ) / cosA =
With the pythagorean identities, cos^2 A + sin^2 A = 1
1 / cosA =
secA = secA
2007-06-11 02:11:27 · answer #3 · answered by Alex 4 · 0⤊ 0⤋
cos A + tan A sin A = cos A + (sin A)^2 / Cos A
= (Cos^2 A + Sin^2 A )/ cos A
= 1/ cos A
= Sec A
2007-06-11 02:13:36 · answer #4 · answered by astrokid 4 · 0⤊ 0⤋
LHS
cos A + (sin A/cos A).sin A
= cos A + sin²A / cos A
= (cos²A + sin²A) / cos A
= 1 / cos A
= sec A = RHS
2007-06-11 09:31:37 · answer #5 · answered by Como 7 · 0⤊ 0⤋
well first you can safely multiply by 1 any term so lets multiply the first term by a form of 1, specifically cosA/cosA.
this gives cos^2(a)/cos(a) + tan(a)sin(a) = sec(a)
then tan(a) = sin(a)/cos(a)
so now we have
cos^2(a)/cos(a) + sin^2(a)/cos(a) = sec(a)
now bring out the 1/cos(a) term leaving
1/cos(a) * (cos^2(a) + sin^2(a)) = sec(a)
using the trig identity of sin^2(a) + cos^2(a) = 1 we get
1/cos(a) = sec(a) Q.E.D.
done.
hope it helps:)
2007-06-11 02:21:44 · answer #6 · answered by Anonymous · 0⤊ 0⤋
we know tanA = sin A/cosA
and secA = 1/cosA
therefore
cosA + sin^2A/cosA = 1/cosA
cos^2A + sin^2A = 1
therefore LH=RH
2007-06-11 02:21:48 · answer #7 · answered by theanswerman 3 · 0⤊ 0⤋
cosA+tanAsinA=secA
LHS
cosA+sinA/cosA * sinA (tanA=sinA/cosA)
(cos^2A+sin^2A)/cosA (Taking LCM)
Now, since cos^2A+sin^2A = 1
1/cosA=secA Proved.
LHS=RHS
2007-06-11 02:23:03 · answer #8 · answered by Robin Hood 1 · 0⤊ 0⤋
you can ignore the A's
cos + (sin/cos)sin =1/cos
cos + sin^2/cos=1/cos
cos^2/cos + sin^2/cos = 1/cos
(cos^2 + sin^2)/cos=1/cos
cos^2 + sin^2 is equal to 1 so
1/cos=1/cos
2007-06-11 02:16:15 · answer #9 · answered by mapacheverde 2 · 0⤊ 0⤋
cosA +tanAsinA = secA
l.h.s.=
cosA+(sinA/cosA)sinA
=cosA+sin^2A/cosA
=(cos^2+sin^2A)/cosA
=1/cosA=secA (since cos^2A+sin^2A=1)
2007-06-11 02:15:20 · answer #10 · answered by Sumita T 3 · 0⤊ 0⤋